# $x^2+35=7^n$ has no natural solutions $\,x,n$ [duplicate]

Prove there does not exist $$x,n\in \Bbb{N}$$ s.t. $$x^2+35=7^n$$.

So I can see that $$x$$ must be even, because an odd plus an odd as even and $$7^n$$ is going to be odd, so if there does exists and $$x$$ it'll be even so. $$(2k)^2+35=7^n, \quad k\in \Bbb{Z}.$$ Then I think the next step would be finding a contradiction in $$(2k)^2=7^n-35$$ but I'm not sure how.

• Won't $x$ have to be divisible by $7$? – Lord Shark the Unknown Nov 28 at 11:02
• Clearly $n\ge 2\,$ so by the Lemma in the dupe $\,7^2\mid x^2+7(5)\iff 7\mid \gcd(x,5)\Rightarrow\!\Leftarrow\ \$ – Bill Dubuque Nov 28 at 14:23
• @The They are both special cases of the Lemma in the linked dupe - see my prior comment. We close such question as abstract dupes in order to avoid having thousands of dupe answers that differ only in specializing some parameters. – Bill Dubuque Nov 28 at 17:14
• The accepted answer here clearly shows that one does not necessarily use the mentioned lemma. Also OP's original title shows "Prove there does not exist $x,n\in N$ s.t. $x^2+35=7^n$", which is very different from the linked one. – Jack Dec 2 at 14:19
• @BillDubuque It seems like somebody edited the question title on the linked duplicate. Also, I generally feel weird about closing something as a duplicate of a closed question. I think a better solution here would be to create a new question about your lemma advertised as an abstract-duplicate FAQ and close this question (and the other one) as duplicates of that one. This will be easier to achieve if a moderator helps with the reopening and reclosing. Since you posted the lemma to that question, maybe you would like to make the abstract duplicate question? – Trevor Gunn Dec 5 at 20:36

It is easy to observe from the equation that $$x$$ must be divisible by $$7$$. Seting $$x = 7k$$ , we get $$49k^2 +35 = 7^{n} \implies 7k^2 +5 = 7^{n-1}$$

But note that $$7k^2$$ and $$7^{n-1}$$ are divisible by $$7$$ but $$5$$ is not , which is a contradiction. Hence there is no possible value of $$x,n \in \mathbb{N}$$

• Technically this is not quite complete since we must discuss the case $n=1$, although that's rather trivial since the LHS is at least $35$. – YiFan Nov 28 at 11:58

We have that

$$x^2=7^n-35=7\cdot(7^{n-1}-5)$$

but

$$7^{n-1}-5 \equiv 2 \mod 7$$